In figure, AB is a potentiometer wire, length 10 m and resistance `2 Omega` with open the balancing length is 5.5 m. However, on closing key K the balancing length reduces to 5 m. The initial resistance of the cell `E_(1)` is

In the givenn figure AB is a potentiometer wire of length 10 m and resistance 1Omega with key K open the balancing length is 5.5 m. however on closing key K the balancing length reduces to 5m. The initial resistance of the cell E_(1) is

In figure AB is a potentiometer wire of length 10 m and resistance 1Omega . The balancing length for 2V potential drop is 18 m. The e.m.f. of the battery E is

In given figure, the potentiometer wire AB has a resistance of 5Omega and length 10 m. The balancing length AM for the e.m.f. of 0.4 V is

In given figure, the potentiometer wire AB has a resistance of 5 Omega and length 10 m . The balancing length AM for the emf of 0.4 V is

In given figure, the potentiometer wire AB has a resistance of 5 Omega and length 10 m . The balancing length AM for the emf of 0.4 V is

In figure,the potentiometer wire A B has a resistance of 5 Omega and length 10m. The balancing length AJ for the cell of e.m.f.of 0.4 V is

A cell of e.m.f. 1.2 V is balanced by 150 cm of potentiometer wire. When the cell is shunted by a resistance of 4 Omega , the balancing length is reduced by 30 cm. What is the internal resistance of the cell ?

In the determination of internal resistance of a cell of e.in.f. 1.5 V with potentiometer, the balancing length is 75 cm in the open circuit and when a resistance of 10 Omega is used the balancing length is found to be 65 cm. Calculate the internal resistance of the cell.

In an experiment to find the internal resistance of a cell by a potentiometer, a balance was obtained for 50 cm length of the potentiometer wire, with a cell of e.m.f. 2V. When the cell was shunted by a resistance of 2 Omega , the balancing length of the potentiometer wire was 40 cm. What was the internal resistance of the cell ?